How did the authors arrive at $$\frac{3}{2}\left(\frac{S_{i j}}{\sigma_{\mathrm{e}}}\right)$$ in the second equation below? Chain rule is obvious but, I can't get the first term.

With the consideration of initial yield stress, $k$, and ignoring the work hardening and other state variables, energy dissipation potential can be in the form of $$ \psi=\frac{K}{n+1}\left(\frac{\sigma_{\mathrm{e}}-k}{K}\right)^{n+1}\tag{1} $$ Where $\sigma_{\mathrm{e}}=\left(3 S_{i j} \cdot S_{i j} / 2\right)^{1 / 2}$ is the effective stress, $S_{i j}=$ $\sigma_{i j}-\delta_{i j} \sigma_{k k} / 3$ is the component of stress deviator tensor (the Einstein summation convention of summing on repeated indices is used in this paper), $\sigma_{i j}$ is the component of stress tensor and $\delta_{i j}$ is the kronecker delta. $K$ and $n$ are material constants. Assuming normality and the associated flow rule, the multiaxial relationship is given by $$ \frac{\mathrm{d} \varepsilon_{i j}^{\mathrm{p}}}{\mathrm{d} t}=\frac{\partial \psi}{\partial S_{i j}}=\frac{3}{2}\left(\frac{S_{i j}}{\sigma_{\mathrm{e}}}\right)\left(\frac{\sigma_{\mathrm{e}}-k}{K}\right)^{n}\tag{2} $$ Where $\varepsilon_{i j}^{\mathrm{p}}$ is the component of plastic strain tensor.

The paper referred here:

Chen, Y., Zhuang, W., Wang, S., Lin, J., Balint, D., & Shan, D. (2012). Investigation of FE model size definition for surface coating application. Chinese Journal of Mechanical Engineering, 25(5), 860-867.



1 Answer 1


Taking into account that $$ \sigma_e=\left(\frac{3}{2}S_{ij}S_{ij}\right)^{1/2} $$ then $$ \frac{\partial\sigma_e}{\partial S_{hk}}= \frac{1}{2}\left(\frac{3}{2}S_{ij}S_{ij}\right)^{-1/2}\frac{\partial}{\partial S_{hk}}\frac{3}{2}S_{ij}S_{ij}= \frac{1}{2\sigma_e}\cdot\frac{3}{2}\frac{\partial}{\partial S_{hk}}S_{ij}S_{ij}= \frac{1}{2\sigma_e}\cdot\frac{3}{2}2S_{hk}= \frac{3}{2}\left(\frac{S_{hk}}{\sigma_e}\right) $$


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